Respuesta :
Answer:
[tex]\dfrac{dy}{dt}=0.27-0.009y(t),$ y(0)=60kg[/tex]
Step-by-step explanation:
Volume of water in the tank = 1000 L
Let y(t) denote the amount of salt in the tank at any time t.
Initially, the tank contains 60 kg of salt, therefore:
y(0)=60 kg
Rate In
A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.
[tex]R_{in}[/tex] =(concentration of salt in inflow)(input rate of solution)
[tex]=(0.03\frac{kg}{liter})( 9\frac{liter}{min})=0.27\frac{kg}{min}[/tex]
Rate Out
The solution is mixed and drains from the tank at the same rate.
Concentration, [tex]C(t)=\dfrac{Amount}{Volume} =\dfrac{y(t)}{1000}[/tex]
[tex]R_{out}[/tex] =(concentration of salt in outflow)(output rate of solution)
[tex]=\dfrac{y(t)}{1000}* 9\dfrac{liter}{min}=0.009y(t)\dfrac{kg}{min}[/tex]
Therefore, the differential equation for the amount of Salt in the Tank at any time t:
[tex]\dfrac{dy}{dt}=R_{in}-R_{out}\\\\\dfrac{dy}{dt}=0.27-0.009y(t),$ y(0)=60kg[/tex]
